Conjecture on tangent-map morphisms for rational curves in projective space

From papers

Let dn+1d\geq n+1 be integers. Set

q=dn,a=(q+1)nd,b=dnq.q=\left\lfloor \frac{d}{n} \right\rfloor,\qquad a=(q+1)n-d,\qquad b=d-nq.

For a morphism

f:OP1(2)OP1(d+q)aOP1(d+q+1)b,f:\mathcal{O}_{\mathbb{P}^1}(2)\longrightarrow \mathcal{O}_{\mathbb{P}^1}(d+q)^{\oplus a}\bigoplus \mathcal{O}_{\mathbb{P}^1}(d+q+1)^{\oplus b},

where CϕPnC\xrightarrow{\phi}\mathbb{P}^n denotes a rational curve of degree dd, the following assertions hold. Tangent-map conjecture. (1) If char(K)=0\operatorname{char}(K)=0, n=2,3n=2,3 or dn+2d\geq n+2, then a general such morphism ff is induced by a rational curve, namely there is a curve CϕPnC\xrightarrow{\phi}\mathbb{P}^n such that f=ϕdϕf=\phi^*\circ d\phi. (2) If n4n\geq4, there exists a morphism

f:OP1(2)OP1(n+1)(n1)OP1(n+3)f:\mathcal{O}_{\mathbb{P}^1}(2)\longrightarrow \mathcal{O}_{\mathbb{P}^1}(n+1)^{\oplus(n-1)}\oplus\mathcal{O}_{\mathbb{P}^1}(n+3)

that is not induced by any rational curve CϕPnC\xrightarrow{\phi}\mathbb{P}^n of degree n+1n+1, so that fdϕϕf\ne d\phi\circ\phi^* for every such curve. The question concerns which morphisms arise from composing the tangent map of a rational curve with restriction to the curve; the case dn+1d\geq n+1 was stated as open before these main results, while the displayed assertions provide the claimed positive and negative answers in the indicated cases.

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Sources & referencesView supporting material

Primary source

Chen Song, “The Syzygy Matrix and the Differential for Rational Curves in Projective Space”, arXiv:2409.03985 (2024).

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